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CHAPTER 19 OPERATIONAL AMPLIFIERS
Exercise 108, Page 307
1. A differential amplifier has an open-loop voltage gain of 150 when the input signals are 3.55 V
and 3.40 V. Determine the output voltage of the amplifier.
From equation (1), output voltage, V o = A o (V 2 - V 1 ) = 150(3.55 – 3.40)
= (150)(0.15) = 22.5 V
2. Calculate the differential voltage gain of an op amp that has a common-mode gain of 6.0 and a
CMRR of 80 dB
 differential voltage gain 
CMRR = 20log10 
 dB
 common mod e gain 
i.e.
 diferential voltage gain 
80 = 20 log10 

6.0


from which,
80
 differential voltage gain 
 log10 

20
6.0


80
Hence,
10 20 
differential voltage gain
6.0
and differential voltage gain = 6.0 10  = 6  104
4
3. A differential amplifier has an open-loop voltage gain of 150 and a common input signal of
4.0 V to both terminals. An output signal of 15 mV results. Determine the common-mode gain
and the CMRR.
Common-mode gain, A com
Vo
15  103


= 0.00375 or 3.75  10 3
Vcom
4.0
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 differential voltage gain 
CMRR = 20log10 
 dB
 common mod e gain 
 150 
= 20 log10 
 dB  20 log10 40000 = 92.04 dB
 0.00375 
4. In the inverting amplifier of shown below, R i = 1.5 k and R f = 2.5 k. Determine the output
voltage when the input voltage is: (a) + 0.6 V (b) - 0.9 V
 R f 
From equation (5), V o = 
 Vi
 Ri 
 2500 
(a) When V i = + 0.4 V, V o = 
 (+ 0.6) = - 1.0 V
 1500 
 2500 
(b) When V i = - 1.2 V, V o = 
 (- 0.9) = + 1.5 V
 1500 
5. The op amp shown below has an input bias current of 90 nA at 20C. Calculate (a) the voltage
gain, and (b) the output offset voltage due to the input bias current.
(a) Voltage gain, A = 
Rf
1.2 106

= - 80
Ri
15 103
 15 103 1.2 106 
 Ri Rf 
9

(b) Offset voltage, Vos  IB 
   90 10  
3
6
R

R
15

10

1.2

10
 
 
 i
f 

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 90 10 18 10  = 1.33 mV
=
9
9
1215000
6. Determine (a) the value of the feedback resistor, and (b) the frequency for an inverting amplifier
to have a voltage gain of 45 dB, a closed-loop bandwidth of 10 kHz and an input resistance of
20 k.
(a) Gain in decibels = 20log10  voltagegain 
i.e.
45 = 20 log10 A
and
A = 10 20 = 177.83
Also,
A=
from which,
45
 log10 A
20
45
Rf
Ri
i.e.
177.83 =
Rf
20  103
from which, feedback resistor, R f  177.83  20 103 = 3.56 M
(b) Frequency = gain  bandwidth = 177.83  10  103 = 1.78 MHz
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247
Exercise 109, Page 313
1. If the input voltage for the op amp shown below is – 0.5 V, determine (a) the voltage gain,
(b) the output voltage.
(a) Voltage gain, A = 1 +
15 103
Rf
=1+
= 1 + 2.206 = 3.206 or 3.21
Ri
6.8 103
 R 
(b) Output voltage, Vo  1  f  Vi  (3.206)(0.5) = - 1.60 V
 Ri 
2. In the circuit shown below, determine the value of the output voltage, Vo , when (a) V1 = + 1 V
and V2 = + 3 V (b) V1 = + 1 V and V2 = - 3 V
V V 
3 
 1
(a) Output voltage, Vo  R f  1  2     25 103  

3
3 
 10 10 10 10 
 R1 R 2 
=   25 103 100 10 6  300 10 6 
=   25 103  400 106  = - 10 V
V V 
3 
 1
(b) Output voltage, Vo  R f  1  2     25 103  

3
3 
 10 10 10 10 
 R1 R 2 
=   25 103 100 10 6  300 10 6 
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=   25  103  200  106  = + 5 V
3. For the summing op amp shown below, determine the output voltage, Vo
V V V 
0.5
0.8 
 0.3
Output voltage, Vo  R f  1  2  3     60 103  


3
3
3 
 15 10 25 10 32 10 
 R1 R 2 R 3 
=   60 103  20 10 6  20 10 6  25 10 6 
=   60 103  65 106  = - 3.9 V
4. A steady voltage of – 1.25 V is applied to an op amp integrator having component values of
R = 125 k and C = 4.0 F. Calculate the value of the output voltage 120 ms after applying the
input, assuming that the initial capacitor charge is zero.
Output voltage, Vo  
1
1
Vi dt  
 1.25 dt

CR
 4.0 106 125 103  
=
1
 1.25  dt  2 1.25 t  = 2.5 t
0.5 
When time t = 120 ms, output voltage, Vo   2.5  120 103  = 0.3 V
5. In the differential amplifier shown below, determine the output voltage, Vo , if: (a) V1 = 4 mV
and V2 = 0 (b) V1 = 0 and V2 = 6 mV (c) V1 = 40 mV and V2 = 30 mV (d) V1 = 25 mV and
V2 = 40 mV
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(a) Output voltage, Vo  
 120 103 
Rf
Vi   
4 103  = - 60 mV
3 
Ri
 8 10 
 R 3  R f
(b) Output voltage, Vo  
1 
 R 2  R3   Ri

 120  120 
3
 V2  
1 
  6 10 
8 
 8  120 

 120 
3
=
 1  15   6 10  = + 90 mV
 128 
 R
(c) V1  V2 hence, output voltage, Vo   V1  V2    f
 Ri

 120 
   40  30   
 mV = - 150 mV
 8 

 R 3  R f 
(d) V2  V1 hence, output voltage, Vo   V2  V1  
1 

R

R
Ri 
3 
 2
 120  120 
=  40  25  
 1 
 = + 225 mV
8 
 128  
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